1. Find $n$: all 5 girls stand consecutively

Treat the 5 girls as one single block.

Then we have:

• 5 boys
• 1 block of 5 girls

So total objects to arrange $= 6$.

Number of ways to arrange these 6 objects: $6!$

Within the girls' block, the 5 girls can permute among themselves in: $5!$ ways.

Hence, $n = 6!\cdot 5!$

2. Find $m$: exactly 4 girls stand consecutively

We need a block of exactly 4 consecutive girls, but not all 5 consecutive.

Step 2.1: Choose the 4 girls forming the consecutive block

This can be done in: $\binom{5}{4} = 5$ ways.

Arrange these 4 girls within the block in: $4!$ ways.

Now we have the following 7 objects:

• 5 boys
• 1 block of 4 girls
• 1 remaining girl

So total objects $= 7$.

If we arrange these freely, number of arrangements is: $7!$

Thus initial count is: $5\cdot 4!\cdot 7!$

Step 2.2: Subtract cases where all 5 girls become consecutive

This happens when the remaining girl stands adjacent to the block of 4 girls.

For any arrangement of the 6 objects consisting of:

• 5 boys
• 1 block of 4 girls

there are $6!$ ways to arrange them.

Now the remaining girl can be attached to the 4-girl block on either side to make all 5 girls consecutive: $2 \text{ ways}$

Also:

• choose the 4 girls in the block: $5$ ways,
• arrange them inside: $4!$ ways.

So number of unwanted arrangements is: $5\cdot 4!\cdot 6!\cdot 2$

Therefore, $m = 5\cdot 4!\left(7! - 2\cdot 6!\right)$

Now simplify: $7! - 2\cdot 6! = 7\cdot 6! - 2\cdot 6! = 5\cdot 6!$

Hence, $m = 5\cdot 4!\cdot 5\cdot 6!$

3. Compute $\dfrac{m}{n}$

We have $n = 6!\cdot 5! = 6!\cdot 5\cdot 4!$

and $m = 25\cdot 4!\cdot 6!$

Thus, $\frac{m}{n} = \frac{25\cdot 4!\cdot 6!}{6!\cdot 5\cdot 4!} = \frac{25}{5} = 5$

4. Final Answer

$\boxed{5}$